1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Direction vector

  1. Oct 18, 2011 #1
    1. The problem statement, all variables and given/known data

    2. The Celsius temperature in a region of space is given by T(x, y, z) = 2x^2 − xyz. A
    particle is moving in this region with its position at time t given by x = 2t^2, y = 3t,
    z = −t^2, where time and distance are measured in seconds and meters, respectively.
    (a) Find the rate of change of the temperature experienced by the particle in C◦/m
    when the particle is at the point P(8, 6,−4).
    (b) Find the rate of change of the temperature experienced by the particle in C◦/sec
    at P.

    2. Relevant equations



    3. The attempt at a solution

    To find part a, we need to find the direction vector. However, how to find direction vector? Is velocity vector the direction vector?

    If so, than part b is very easy.
    Since [itex]\frac{dT}{dt}[/itex]=[itex]\frac{dT}{ds}\frac{ds}{dt}[/itex]
    Where [/itex]=[itex]\frac{ds}{dt}[/itex] is the magnitute of velocity.

    I wonder whether I am correct.
     
  2. jcsd
  3. Oct 18, 2011 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The "direction" vector is the unit tangent vector to the path. The "rate of change of T in degrees Celcius per meter" is [itex]\nabla T\cdot \vec{v}\text{ }[/itex] where [itex]\vec{v}\text{ }[/itex] is the unit tangent vector.

    For part (b) at least, you do NOT need to find the "direction" or s, just use the chain rule:
    [tex]\frac{dT}{dt}= \frac{\partial T}{\partial x}\frac{dx}{dt}+ \frac{\partial T}{\partial u}\frac{dy}{dt}+ \frac{\partial T}{\partial z}\frac{dz}{dt}[/tex].

    That is almost the same as (a) except that you are not dividing by the length of the tangent vector to get the unit tangent vector.
     
  4. Oct 18, 2011 #3
    That means we have 2 ways to do part b?
    1. Is the chain rule.
    2. Just like part (a) but without dividing the length of the tangent vector?
     
  5. Oct 18, 2011 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, the derivative of < x, y, z> with respect to t gives a "tangent vector", dividing by its length gives the unit tangent vector. The chain rule essentiall is the dot product of grad T with that non-unit tangent vector.

    Probably the simplest way to do this is to do (b) first, by the chain rule, then divide that answer by the length of the vector
    [tex](dx/dt)\vec{i}+ (dy/dt)\vec{j}+ (dz/dt)\vec{j}[/tex]
    to get the answer to (a). That is the magnitude of the velocity vector and so is the speed with which the particle is moving. "rate of change of T as a function of time" divided by "rate of change of position as a function of time" is "rate of change of T as a function of distance. In a very loose notation:
    [tex]\frac{\frac{dT}{dt}}{\frac{dx}{dt}}= \frac{dT}{dt}\frac{dt}{dx}= \frac{dT}{dx}[/tex]
     
    Last edited: Oct 18, 2011
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Direction vector
  1. Direction Vector (Replies: 5)

  2. Direct Sum of vectors (Replies: 3)

Loading...